Sublimation of a naphthalene rod in cross-flow air: 23-Chem-A3 Heat and Mass Transfer, May 2015, Q3

Question 3: Sublimation of a naphthalene rod in cross-flow air (Part B — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A slender cylinder sublimes into an air stream flowing perpendicular to its axis. The forced-convection external mass-transfer coefficient (Hilpert cylinder correlation) sets the sublimation rate; the surface vapour concentration is fixed by the naphthalene vapour pressure.

Quantity	Value
Rod diameter $d$	3 mm $=0.30$ cm
Air velocity $v$	91.44 cm/s
Temperature $T$	34 °C $=307$ K
Diffusivity $D_{AB}$	$8\times10^{-6}$ m²/s $=0.08$ cm²/s
Vapour pressure $p^{\text{vap}}$	26 Pa
Air kinematic viscosity $\nu$	0.16 cm²/s
MW / density	128 / 1.15 g/cm³

Find. The time for the rod to lose 10% of its mass.

Check — data note

The printed diffusivity “$8\times10^{-6}$ cm²/s” is physically impossible for a small gas molecule in air (it would give $Sc\approx2\times10^{4}$, a liquid-like value). The intended value is $8\times10^{-6}$ m²/s $=0.08$ cm²/s — the well-known naphthalene–air diffusivity, giving $Sc\approx2$. We solve with $D_{AB}=0.08$ cm²/s; the exam’s “state any assumptions” clause covers this correction.

Figure 3 — Cross-flow sublimation of a cylinder. The Hilpert analogy $\overline{Sh}=0.683\,Re^{0.466}Sc^{1/3}$ (valid $Re\approx40$–4000) gives the external coefficient; the surface is saturated at the naphthalene vapour pressure while the bulk air is clean.

Approach. Form $Re$ and $Sc$, apply the Hilpert cross-flow-cylinder analogy for $\overline{Sh}$, convert to $k_c$, evaluate the saturated surface concentration $c_{As}=p^{\text{vap}}/RT$, and divide 10% of the rod’s mass (per unit length) by the sublimation rate (per unit length).

Reynolds and Schmidt numbers. $$Re=\frac{vd}{\nu}=\frac{(91.44)(0.30)}{0.16}=171,\qquad Sc=\frac{\nu}{D_{AB}}=\frac{0.16}{0.08}=2.0.$$
Sherwood number (Hilpert cylinder analogy). For $Re\approx40$–4000, $$\overline{Sh}=0.683\,Re^{0.466}Sc^{1/3}=0.683\,(171)^{0.466}(2.0)^{1/3}=9.46.$$ (The Churchill–Bernstein correlation agrees to within 5%.)
Mass-transfer coefficient. $$k_c=\frac{\overline{Sh}\,D_{AB}}{d}=\frac{(9.46)(0.08)}{0.30}=2.52\ \text{cm/s}.$$
Surface vapour concentration. Saturated at the surface, $$c_{As}=\frac{p^{\text{vap}}}{RT}=\frac{26}{(8.314)(307)}=1.02\times10^{-2}\ \text{mol/m}^3=1.02\times10^{-8}\ \text{mol/cm}^3.$$
Sublimation rate per unit length. Mass flux $\dot m''=k_c\,c_{As}\,M=(2.52)(1.02\times10^{-8})(128)=3.29\times10^{-6}$ g/cm²·s; times the perimeter $\pi d$: $$\dot m'=\dot m''\,\pi d=(3.29\times10^{-6})(\pi)(0.30)=3.10\times10^{-6}\ \text{g/cm}\cdot\text{s}.$$
Time for 10% mass loss. Mass per unit length $m'=\rho\,\tfrac{\pi}{4}d^2=(1.15)(\tfrac{\pi}{4})(0.30)^2=0.0813$ g/cm, so $$t=\frac{0.10\,m'}{\dot m'}=\frac{0.10(0.0813)}{3.10\times10^{-6}}=2.62\times10^{3}\ \text{s}=\boxed{44\ \text{min}}.$$

Quantity	Result
$Re$ / $Sc$	171 / 2.0
$\overline{Sh}$ / $k_c$	9.46 / 2.52 cm/s
Surface concentration $c_{As}$	$1.02\times10^{-8}$ mol/cm³
Time for 10% loss	$2.6\times10^{3}$ s ≈ 44 min

Check

The rod loses only 10% of its mass, so its diameter shrinks by just $1-\sqrt{0.9}=5\%$ ($d\propto\sqrt{m}$); treating $Re$, $Sh$ and the area as constant introduces negligible error, which justifies the single-step (rather than integrated-shrinkage) calculation.

Concept & Theory

Topic. Forced-convection sublimation from a cylinder in cross-flow.

Key relations. $Re=vd/\nu$; $Sc=\nu/D_{AB}$; Hilpert analogy $\overline{Sh}=0.683\,Re^{0.466}Sc^{1/3}$ (Re 40–4000); $k_c=\overline{Sh}\,D_{AB}/d$; saturated surface $c_{As}=p^{\text{vap}}/RT$; rate $\dot m'=k_c c_{As}M\pi d$.

Why this works. The Chilton–Colburn heat–mass analogy lets the Hilpert cylinder Nusselt correlation be reused for Sherwood by swapping $Pr\to Sc$. Because naphthalene is only sparingly volatile ($p^{\text{vap}}=26$ Pa $\ll P$), the surface concentration is dilute, the bulk air is essentially clean, and the low-mass-flux (dilute) form of the flux law applies without a Stefan correction.

Common pitfall. Taking the printed diffusivity at face value (unit error — see the data note), or forgetting that $c_{As}$ uses the surface temperature and the naphthalene vapour pressure, not the air composition. Watch the cgs bookkeeping: the perimeter is $\pi d$, the cross-section $\tfrac{\pi}{4}d^2$.

Source. Welty et al., §30 (flow past a single cylinder, mass transfer); Geankoplis, §7.3 (external forced-convection correlations); Incropera & DeWitt (Hilpert constants).

Question 3 of 7: Sublimation of a naphthalene rod in cross-flow air

Question 3: Sublimation of a naphthalene rod in cross-flow air (Part B — equal value)